2004 AIME I Problems/Problem 6
We divide the problem into two cases: one in which zero is one of the digits and one in which it is not. In the latter case, suppose we pick digits such that . There are five arrangements of these digits that satisfy the condition of being snakelike: , , , , . Thus there are snakelike numbers which do not contain the digit zero.
In the second case we choose zero and three other digits such that . There are three arrangements of these digits that satisfy the condition of being snakelike: , , . Because we know that zero is a digit, there are snakelike numbers which contain the digit zero. Thus there are snakelike numbers.
Let's create the snakelike number from digits , and, if we already picked the digits there are 5 ways to do so, as said in the first solution. And, let's just pick the digits from 0-9. This get's a total count of But, this over-counts since it counts numbers like 0213. We can correct for this over-counting. Lock the first digit as 0 and permute 3 other chosen digits . There are 2 ways to permute to make the number snakelike, b-a-c, or c-a-b. And, we pick a,b,c from 1 to 9, since 0 has already been chosen as one of the digits. So, the amount we have overcounted by is . Thus our answer is
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